Parametric polynomial solution of a single polynomial equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:22:06Z http://mathoverflow.net/feeds/question/11336 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11336/parametric-polynomial-solution-of-a-single-polynomial-equation Parametric polynomial solution of a single polynomial equation Ewan Delanoy 2010-01-10T15:41:09Z 2010-01-10T16:09:13Z <p>Let $P$ be a polynomial in $n$ variables with rational coefficients, $P \in {\mathbb Q}[Z_1,Z_2, \ldots ,Z_n]$, and consider the algebraic set<br /> $Z=\lbrace (z_1,z_2,z_3, \ldots ,z_n) \in {\mathbb Q}^n | \ P(z_1,z_2, \ldots ,z_n)=0 \rbrace$ </p> <p>If $r$ is a nonnegative integer, $x_1,x_2, \ldots ,x_r$ are variables, and $Q_1,Q_2, \ldots ,Q_n$ are polynomials in $x_1,x_2, \ldots ,x_r$ such that $(Q_1(x_1, \ldots ,x_r),Q_2(x_1, \ldots ,x_r),\ldots,Q_n(x_1, \ldots ,x_r)) \in Z$ for all $(x_1, \ldots ,x_r) \in {\mathbb Q}^r$, we call $(Q_1,Q_2, \ldots ,Q_n)$ a $r$-dimensional parametric solution of the equation $P(z_1,z_2, \ldots ,z_n)=0$. It is also natural to define a maximal parametric solution as one with the largest possible $r$ (to avoid trivialties, we also impose that there is no variable upon which none of the $Q_i$ depends. I'm not sure that this last condition avoids all degenerate cases, but I'd like to avoid definitions that involve advanced notions such as the dimension of an algebraic variety ).</p> <p>My questions : is the problem of computing the largest $r$ known to be undecidable in general ? What are the most general cases in which algebraic geometry allows us to compute the largest $r$ (and the associated parametric solutions) ?</p> http://mathoverflow.net/questions/11336/parametric-polynomial-solution-of-a-single-polynomial-equation/11344#11344 Answer by Felipe Voloch for Parametric polynomial solution of a single polynomial equation Felipe Voloch 2010-01-10T16:09:13Z 2010-01-10T16:09:13Z <p>I think the problem of deciding if $r=0$ or $r > 0$ will be related to (but a bit more subtle than) Hilbert's 10th problem over $Q(t)$ so it might already be undecidable. I haven't seen a statement in this direction but I am not an expert on undecidability questions.</p> <p>For curves, $r=0$ or $1$ and $r=1$ if and only if $Z$ has genus zero and a rational point and this is decidable.</p> <p>For surfaces, maybe you can decide if $r=2$ or not. It's a matter of deciding if the surface is a rational surface. This can certainly be done over an algebraically closed field but I'm not sure what happens over the rationals. Now, to decide if $r=0$ or $1$ in general, you need to be able to decide if a surface contains rational curves and I believe we don't know how to do that for arbitrary surfaces. Although we do know how to decide if a surface fibered over a rational curve has a section, which would be the situation of Hilbert's 10th problem over $Q(t)$.</p>